Description：
Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.
Calling next() will return the next smallest number in the BST.
Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
实现一个迭代器来遍历二叉查找树。当然首先想到的方法就是中序遍历把有序元素保存在容器中，顺序操作。但是要求uses O(h) memory就不能这样干了。
这里用一个栈来做暂存器，先把所有左节点压栈，这时栈定就一定是最小的元素。之后出栈返回当前的栈顶元素，对于栈顶元素，把以其为顶点的右子树的所有左节点都压栈。以此类推，直到所有元素都遍历一遍。在uses O(h) memory的要求下完成了二叉查找树的遍历。
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class BSTIterator {
public Stack<TreeNode> stack;
public BSTIterator(TreeNode root) {
stack = new Stack<TreeNode>();
while(root != null) {
stack.push(root);
root = root.left;
}
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.empty();
}
/** @return the next smallest number */
public int next() {
TreeNode node = stack.peek();
int nextVal = node.val;
stack.pop();
TreeNode t = node.right;
while(t != null) {
stack.push(t);
t = t.left;
}
return nextVal;
}
}
/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/
LeetCode——Binary Search Tree Iterator
原文：http://www.cnblogs.com/wxisme/p/4606401.html